AP EAMCET · Maths · Definite Integration
For \(n \in N\), if \(I_n=\int \frac{\sin n x}{\sin x} d x=\frac{2}{n-1} \sin (n-1) x+I_{n-2}\) and \(\int_0^\pi \frac{\sin n x}{\sin x} d x=\frac{k \pi}{2}\), then \(k=\)
- A \((-1)^n-1\)
- B \(1-(-1)^n\)
- C \((-1)^n\)
- D \((-1)^{n+1}\)
Answer & Solution
Correct Answer
(B) \(1-(-1)^n\)
Step-by-step Solution
Detailed explanation
\(\begin{aligned} & \text { Consider } I_n=\int_0^\pi \frac{\sin n x}{\sin x} d x, n \in N \\ & I_n=\left[\frac{2}{n-1} \sin (n-1) x\right]_0^n+I_{n-2}=I_{n-2}\end{aligned}\)…
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